Question

64. In the reaction above, hydrogen carbonate is:
                an acid a base   a reducing agent  or an oxidizing agent

65. How many moles of lithium hydroxide would you need to make 0.10 mol H₂O? (you may assume that you have excess carbonic acid)

b. Given 300. g of lithium hydroxide 300. g carbonic acid, what is the limiting reagent?

c. Given 300. g of lithium hydroxide 300. g carbonic acid, what is the theoretical yield of lithium carbonate in g?

Answer 1

65. 2 LiOH + H2CO3 ------------- Li2CO3 + 2 H2O

       2 mole                                              2mole

According to equation 2 mole of H2O = 2 mole of LiOH

                                0.10 mole of H2O = ?

                                                               = 0.10 x 2/2 = 0.10mole of LiOH

Number of moles LiOH = 0.10mole

b) mass of LiOH = 300.0 grams

Molar mass of LiOH= 23.95 grams/mol

Number of moles of LiOH = 300.0/23.95 = 12.53 moles

Mass of H2CO3 = 300.0 grams

Molar mass of H2CO3 = 62.03 gram/mole

number of moles of H2CO3 = 300.0/62.03 = 4.84 moles

According to equation,

1mole of H2CO3 = 2 mole of LiOH

4.84 moles of H2CO3 = ?

                         = 2x4.84/1 = 9.68 mole of LiOH

we need 9.68 mole of LiOH.but we have 12.53 moles of LiOH. so it is excess reagent.

Hence H2CO3 is limiting reagent.

c) Accoriding to equation,

1 mole of H2CO3 = 1 mole of Li2CO3

4.84 moles of H2CO3 = ?

                               = 1x4.84/1 = 4.84 moles of L2CO3

number of moles of Li2CO3 = 4.84 moles

Molar mass of Li2CO3 = 73.89 gram/mole

mass of 4.84 moles of Li2CO3 = 4.84 x 73.89 = 357.63 grams

Theoritical yield of Li2CO3 = 357.63 grams.